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A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.
Its volume is then reduced to the original value from $E$ to $F$ by an isobaric process. Calculate the total work done by the gas from $D$ to $E$ to $F$.
Its volume is then reduced to the original value from $E$ to $F$ by an isobaric process. Calculate the total work done by the gas from $D$ to $E$ to $F$.
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Verified Answer
Work done from $D$ to $E$
$$
\begin{aligned}
&=W_{D E}=\frac{1}{2} \times(5-2)(600-300)+(5-2)(300) \\
&=450+900=1350 \mathrm{~J}
\end{aligned}
$$
Work done from $E$ to $F$
$$
=W_{E F}=300(2-5)=-900 \mathrm{~J}
$$
$\therefore$ Work done from $\mathrm{D}$ to $\mathrm{E}$ to $\mathrm{F}$
$=W_{D E F}$
$=W_{D E}+W_{E F}$
$=(1350-900) \mathrm{J}=450 \mathrm{~J}$
$$
\begin{aligned}
&=W_{D E}=\frac{1}{2} \times(5-2)(600-300)+(5-2)(300) \\
&=450+900=1350 \mathrm{~J}
\end{aligned}
$$
Work done from $E$ to $F$
$$
=W_{E F}=300(2-5)=-900 \mathrm{~J}
$$
$\therefore$ Work done from $\mathrm{D}$ to $\mathrm{E}$ to $\mathrm{F}$
$=W_{D E F}$
$=W_{D E}+W_{E F}$
$=(1350-900) \mathrm{J}=450 \mathrm{~J}$
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